更新1:
Prove from first principles
How to prove sin(2x)=2sin(x)cos(x)?
回答 (10)
2006-05-25 8:50 am
✔ 最佳答案
sin(2x) = 2sinxcosx is a trigonometric identity called the double angle identity. It was derived from another identity:sin(x + y) = sinxcosy + cosxsiny
here x and y are two different angles but what if x = y
replace y with x to make all the equations in terms of x, thus the equation becomes:
sin(x + x) = sinxcosx + cosxsinx
note that sinxcosx is also equal to cosxsinx, so simplifying the equation it becomes:
sin(2x) = 2sinxcosx or sin(2x) = 2cosxsinx
2006-06-07 7:40 am
The area of a triangle is 1/2*a*b*sin(C)
Construct a right Triangle(1) with sides Opp, Adj, and Hyp with angle (x). Area(1) = 1/2*Hyp*Adj*sin(x)
Using the adjacent side(Adj) as the line of symmetry, reflect the first triangle to get Triangle(2) with sides Hyp, Hyp, and (2*Opp) and angle(2*x)
Area(2) = 1/2*Hyp*Hyp*sin(2*x)
Area(2) = 2*Area(1)
1/2*Hyp*Hyp*sin(2*x) =2*1/2*Hyp*Adj*sin(x)
solving for sin(2*x) gives
sin(2*x) = 2*sin(x)*Adj/Hyp
Adj/Hyp = cos(x)
therefore,
sin(2*x) =2*sin(x)*cos(x)
Construct a right Triangle(1) with sides Opp, Adj, and Hyp with angle (x). Area(1) = 1/2*Hyp*Adj*sin(x)
Using the adjacent side(Adj) as the line of symmetry, reflect the first triangle to get Triangle(2) with sides Hyp, Hyp, and (2*Opp) and angle(2*x)
Area(2) = 1/2*Hyp*Hyp*sin(2*x)
Area(2) = 2*Area(1)
1/2*Hyp*Hyp*sin(2*x) =2*1/2*Hyp*Adj*sin(x)
solving for sin(2*x) gives
sin(2*x) = 2*sin(x)*Adj/Hyp
Adj/Hyp = cos(x)
therefore,
sin(2*x) =2*sin(x)*cos(x)
2016-11-14 1:15 pm
between the Sum and distinction Identities is sin(a + b) = sin(a)cos(b) + cos(a)sin(b). in the given, sin(2X), the a and the b at the instant are not explicitly stated. yet, sin(2X) could properly be expressed as sin(X + X). for that reason, giving us the form sin(a + b) with a = b = X. Substituting X for the a's and b's in the equation sin(a + b) = sin(a)cos(b) + cos(a)sin(b), the equation sin(X + X) = sin(X)cos(X) + cos(X)sin(X). utilising the Closure sources of Equality: sin(X + X) = sin(X)cos(X) + cos(X)sin(X) sin(2X) = sin(X)cos(X) + sin(X)cos(X) sin(2X) = 2sin(X)cos(X) i'm hoping you will comprehend my answer.. thank you.
2006-06-07 6:42 pm
the general formula is:
Sin(A+B)=SinA.CosB + CosA.SinB ......................... (i)
Sin(2x) can be written as Sin(x+x)
now in equation (i) replace the A and B by x everywhere
you will get
=Sin(x).Cos(x) + Cos(x).Sin(x)
=2Sin(x).Cos(x)
Sin(A+B)=SinA.CosB + CosA.SinB ......................... (i)
Sin(2x) can be written as Sin(x+x)
now in equation (i) replace the A and B by x everywhere
you will get
=Sin(x).Cos(x) + Cos(x).Sin(x)
=2Sin(x).Cos(x)
2006-05-25 9:06 am
we know that sin(a+b)=sina cosb +cosa sinb
now put a=b=x,
sin(x+x)=sinx cosx+cosx sinx=2 sinx cosx
now put a=b=x,
sin(x+x)=sinx cosx+cosx sinx=2 sinx cosx
2006-05-25 8:44 am
As we know that sin (x+x)= sinx cosx +cosx sinx
= 2 sinx cosx
= 2 sinx cosx
2006-05-25 8:43 am
Sin(A+B)=SinA.CosB + CosA.SinB ---general formula
Sin(2x)
=Sin(x+x) (apply general formula here)
=Sin(x).Cos(x) + Cos(x).Sin(x)
=Sin(x).Cos(x) + Sin(x).Cos(x)
=2Sin(x).Cos(x)
Sin(2x)
=Sin(x+x) (apply general formula here)
=Sin(x).Cos(x) + Cos(x).Sin(x)
=Sin(x).Cos(x) + Sin(x).Cos(x)
=2Sin(x).Cos(x)
2006-05-25 8:42 am
We know : Sin(A+B) = SinACosB + CosASinB
Put A=B=x
So , Sin(X+X) = Sin2x = sinxcosx + cosxsinx = 2sinxcosx
Put A=B=x
So , Sin(X+X) = Sin2x = sinxcosx + cosxsinx = 2sinxcosx
2006-05-25 8:41 am
Use the sum rule.
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
In this case,
sin(2x) = sin(x+x)
sin(x+x) = sin(x)cos(x) + cos(x)sin(x)
sin(x+x) = sin(x)cos(x) + sin(x)cos(x)
sin(x+x) = 2sin(x)cos(x)
sin(2x) = 2sin(x)cos(x)
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
In this case,
sin(2x) = sin(x+x)
sin(x+x) = sin(x)cos(x) + cos(x)sin(x)
sin(x+x) = sin(x)cos(x) + sin(x)cos(x)
sin(x+x) = 2sin(x)cos(x)
sin(2x) = 2sin(x)cos(x)
2006-05-25 8:40 am
sin 2x = Sin(x + x) = sin x cos x + sin x cos x = 2 sin x cos x
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